Question: Find $\lim_{h\to 0}\dfrac{8(3+h)^3-8(3)^3}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $8$ (Choice B) B $72$ (Choice C) C $216$ (Choice D) D The limit doesn't exist
Explanation: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $8(3+h)^3-8(3)^3$, we can tell that the function is $f(x)=8x^3$ and the $x$ -value is $3$. In other words, the limit expression is equal to $f'(3)$ for $f(x)=8x^3$. Let's find $f'(x)$ : $f'(x)=8\cdot3x^2=24x^2$ Now let's evaluate $f'(3)$ : $f'(3)=24(3)^2=216$ In conclusion, $\lim_{h\to 0}\dfrac{8(3+h)^3-8(3)^3}{h}=216$.